How can tan 4x be simplified or sec 2x?

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How can tan 4x be simplified or sec 2x?

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Answer 1 · 2015-08-07T16:28:08

Let's say we looked at $tan 4 x$ $tan4x$ . We can use the following identities:

$tan 4 x = \frac{sin 4 x}{cos 4 x}$ $tan4x = (sin4x)/(cos4x)$

$sin 2 x = 2 sin x cos x$ $sin2x = 2sinxcosx$
$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x = cos^2x - sin^2x$

$\Rightarrow \frac{2 sin 2 x cos 2 x}{{cos}^{2} 2 x - {sin}^{2} 2 x}$ $=> (2sin2xcos2x)/(cos^2 2x - sin^2 2x)$

$= \frac{4 sin x cos x ({cos}^{2} x - {sin}^{2} x)}{{({cos}^{2} x - {sin}^{2} x)}^{2} - {(2 sin x cos x)}^{2}}$ $= (4sinxcosx(cos^2x - sin^2x))/((cos^2x - sin^2x)^2 - (2sinxcosx)^2)$

$= \frac{4 sin x cos x ({cos}^{2} x - {sin}^{2} x)}{{({cos}^{2} x - {sin}^{2} x)}^{2} - 4 {sin}^{2} x {cos}^{2} x}$ $= (4sinxcosx(cos^2x - sin^2x))/((cos^2x - sin^2x)^2 - 4sin^2xcos^2x)$

$= \frac{4 sin x cos x (1 - 2 {sin}^{2} x)}{{(1 - 2 {sin}^{2} x)}^{2} - 4 {sin}^{2} x {cos}^{2} x}$ $= color(blue)((4sinxcosx(1 - 2sin^2x))/((1 - 2sin^2x)^2 - 4sin^2xcos^2x))$

I don't know if you can get it any simpler; it's all $sin x$ $sinx$ and $cos x$ $cosx$ now, though.

You could also have used:
$tan (2 x + 2 x) = \frac{tan (2 x) + tan (2 x)}{1 - tan (2 x) tan (2 x)}$ $tan(2x+2x) = (tan(2x)+tan(2x))/(1-tan(2x)tan(2x))$

$= \frac{2 tan (2 x)}{1 - {tan}^{2} (2 x)}$ $= (2tan(2x))/(1-tan^2(2x))$

but that's gonna be uglier to simplify (unless you stop here).

$sec (2 x)$ $sec(2x)$ is much simpler.

$= \frac{1}{cos (2 x)} = \frac{1}{{cos}^{2} x - {sin}^{2} x} = \frac{1}{1 - 2 {sin}^{2} x}$ $= 1/(cos(2x)) = 1/(cos^2x - sin^2x) = color(blue)(1/(1-2sin^2x))$

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How can tan 4x be simplified or sec 2x?

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